Mathematics 9709/61 — October/November 2019

Walkthrough

Shona has two possible ways to get to college: bus or cycle. These two ways are mutually exclusive, so the total probability of her not being late is the sum of the probability that she takes the bus and is not late plus the probability that she cycles and is not late.

For each way, multiply the probability of choosing that way by the conditional probability of not being late given that way. If she catches the bus, the chance she is late is $0.4$, so the chance she is not late is $0.6$. If she cycles, the chance she is late is $x$, so the chance she is not late is $1-x$.

This gives the equation $0.8 \times 0.6 + 0.2(1-x) = 0.63$. Expand and solve: $0.48 + 0.2 - 0.2x = 0.63$, so $0.68 - 0.2x = 0.63$, giving $0.2x = 0.05$ and $x = 0.25$.

Key Takeaways

This question uses the law of total probability: if events partition the sample space, the overall probability is the sum of the probabilities of each branch. It also uses complementary probabilities: the probability of not being late is $1$ minus the probability of being late for each mode of travel.

Common Mistakes

• Using $0.4$ as the probability of not being late when catching the bus. The complement must be used: $1 - 0.4 = 0.6$.
• Forgetting that if the probability of being late when cycling is $x$, then the probability of not being late when cycling is $1-x$.
• Making a sign error when expanding $0.2(1-x)$; it becomes $0.2 - 0.2x$, not $0.2 + 0.2x$.
• Not writing the equation $0.8 \times 0.6 + 0.2(1-x) = 0.63$; the mark scheme requires the equation to be shown for method marks.

Things to Be Careful About

• The value of $x$ must be a probability, so it must lie between $0$ and $1$. Here $x = 0.25$ is valid.
• Check which probability the question gives: it is the probability of not being late ($0.63$), so the equation should use the not-late branches, not the late branches.
• An alternative valid equation uses the late branches: $0.8 \times 0.4 + 0.2x = 1 - 0.63 = 0.37$. Both approaches lead to the same answer.